Question

The rise in the water level in a capillary tube of radius 0.07 cm when dipped vertically in a beaker containing water of surface tension 0.07 \[N{m^{ - 1}}\] is $(g = 10m{s^{ - 2}})$
(A) 2 cm
(B) 4 cm
(C) 1.5 cm
(D) 3 cm

Answer 1

Hint When a capillary tube is inserted in water or any other liquid. It experiences capillary rise or fall depending on the density of the liquid. We know that the rise in height of the capillary is directly proportional to surface tension and angle of contact. Also, it is inversely proportional to the radius of the tube, the density of the liquid, and the acceleration due to gravity.

Complete step-by-step answer:
We are given that the radius of the capillary tube is, $r = 0.07cm = 0.07 \times {10^{ - 2}}m$
The surface tension of water is also given, $S = 0.07N{m^{ - 1}}$ .
We know that the density of water is, $\rho = {10^3}kg{m^{ - 3}}$ .
The angle of contact for water is zero degrees.
Using the expression for the rise in the capillary tube : $h = \dfrac{{2S\cos \theta }}{{r\rho g}}$
Where h is the rise of a liquid in a capillary tube
S is the surface tension of the liquid used in the capillary tube
 $\theta $ is the angle of contact
r is the radius of the capillary tube
 $\rho $ is the density of the liquid
and g is the acceleration due to gravity
Now, substituting all the given values in the expression for h, we get
 $
   \Rightarrow h = \dfrac{{2 \times 0.07 \times \cos 0^\circ }}{{0.07 \times {{10}^{ - 2}} \times {{10}^3} \times 10}} \\
   \Rightarrow h = 2 \times {10^{ - 2}}m \\
   \Rightarrow h = 2cm \\
 $

Therefore, option (A) is correct.

Note The angle of contact is the angle that a perpendicular to the walls of the capillary makes with the meniscus of the liquid at the point of contact.


This is a diagram of two liquids where liquid(a) has a convex meniscus $(\theta > 90^\circ )$ and liquid(b) has a concave meniscus $(\theta > 90^\circ )$. For water, the meniscus is flat. Therefore, the angle of contact is zero for water.